(i) \(\begin{aligned} & \therefore B=\frac{\mu_0}{4 \pi} \frac{2 M}{d^3} \end{aligned}\)
\(\begin{aligned} \Rightarrow B=\frac{4 \pi \times 10^{-7} \times 2 \times 0.48}{4 \pi \times(0.1)^3}=0.96 \times 10^{-4} \mathrm{~T} \end{aligned}\)
\(\Rightarrow\) B = 0.96 G
The magnetic field is along the S-N direction.
(ii) \(\therefore B=\frac{\mu_0}{4 \pi} \times \frac{M}{d^3}=\frac{4 \pi \times 10^{-7} \times 0.48}{4 \pi \times(0.1)^3}=0.48 \mathrm{G}\)
The magnetic field is along the N-S direction.

Since, magnetic flux increases, value of magnetic flux when the loop moves into a uniform magnetic field. So, the induced current should oppose this increased the value of magnetic flux. Thus, the flow will be from QPSRQ, i.e. anti-clockwise.

The current carrying loop is equivalent to a magnetic dipole. The magnetic dipole does not experience any net force in a uniform magnetic field.

(i) At axial points, the electric potential of a dipole has maximum positive or negative value.
(ii) At equatorial points, the electric potential of a dipole is zero.

\(V_L=V_C ; X_L=X_C ; Z=R ; \cos \phi=\frac{R}{Z}=1\)

As we know \( V_{A C} =-\int_{A}^{C} \vec{E} \cdot \overrightarrow{d l}=-E \int_{A}^{C} d l \cos \theta \)
\(=-5 \times 10^{5} \times \frac{5}{100} \times \frac{4}{5} \)
\( =-20 \times 10^{3} \mathrm{~V} \ \left(\because \int_{A}^{C} d l=A C=5 \mathrm{~cm}\right) \)

1.5\(\Omega\)

(i) Principle
An AC generator is based on the phenomenon of electromagnetic induction which states that whenever magnetic flux linked with a conductor (or coil) changes, an emf is induced in the coil.

Working
As the armature of coil is rotated in the uniform magnetic field, angle \( \theta\) between the.field and normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes and an emf is induced in the coil. According to Fleming's right hand rule, current induced in AB is from A to B and it is from C to D in CD. In the external
circuit, current flows from B₂ to B₁. To calculate the magnitude of emf induced, suppose
A = area of each turn of the coil,
N = number of turns in the coil,
B = strength of magnetic field
and \( \theta\) = angle which normal to the coil makes with B at any instant t.

(ii) As the earth's magnetic field lines are cut by the falling rod, the change in magnetic flux takes place. This change in flux induces an emf across the ends of the rod.
Since, the rod is falling under gravity.
v = gt \([\because u=0]\)
Induced emf, \(\varepsilon=B l v \Rightarrow \varepsilon=B \lg t\)
\(\varepsilon \propto t \)

When the current through the element is very small, heating effects can be ignored and the temperature T₁ of the element is the same as room temperature. When the toaster is connected to the supply, its initial current will be slightly higher than its steady value of 2.68 A. But due to heating effect of the current, the temperature will rise. This will cause an increase in resistance and a slight decrease in current. In a few seconds, a steady state will be reached when temperature will rise no further, and both the resistance of the element and the current drawn will achieve steady values. The resistance R₂ at the steady temperature T₂ is
\(R_{2}=\frac{230 \mathrm{~V}}{2.68 \mathrm{~A}}=85.8 \Omega\)
Using the relation
R₂ = R₁ [1 + α (T₂ - T₁)]
with α = 1.70 x 10^-4 °C^-1, we get
\(T_{2}-T_{1}=\frac{(85.8-75.3)}{(75.3) \times 1.70 \times 10^{-4}}=820^{\circ} \mathrm{C}\)
that is, T₂ = (820 + 27.0) °C = 847 °C
Thus, the steady temperature of the heating element (when heating effect due to the current equals heat loss to the surroundings) is 847 °C.

The capacitive reactance is
\(X_{C}=\frac{1}{2 \pi v C}=\frac{1}{2 \pi(50 \mathrm{~Hz})\left(15.0 \times 10^{-6} \mathrm{~F}\right)}=212 \Omega\)
The rms current is
\(I=\frac{V}{X_{C}}=\frac{220 \mathrm{~V}}{212 \Omega}=1.04 \mathrm{~A}\)
The peak current is
\(i_{m}=\sqrt{2} I=(1.41)(1.04 A)=1.47 A\)
This current oscillates between +1.47A and -1.47 A, and is ahead of the voltage by π/2.
If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Magnitude of one of the magnetic fields, B₁ = 1.2 x 10⁻² T
Angle between the two fields, θ = 60°
At stable equilibrium, the angle between the dipole and field B₁, θ₁ = 15°
Let the magnitude of the other the other magnetic field be B₂
Angle between the dipole and field B₂, θ₂ = θ − θ₁ = 60° − 15° = 45°
At rotational equilibrium, the torques due to both the fields must balance each other. That is,
Torque due to field B1 = Torque due to field B2Torque due to field B1 = Torque due to field B2
⇒ MB1 sin θ₁ = MB2  sin θ2 
\(\Rightarrow \mathrm{B}_{2}=\frac{\mathrm{B}_{1} \times \sin \theta_{1}}{\sin \theta_{2}}\)
Where,
M = Magnetic moment of the dipole
Putting the values, we get,
\(\mathrm{B}_{2}=\frac{1.2 \times 10^{-2} \times \sin 15^{\circ}}{\sin 45^{\circ}}\)
= 4.39 x 10^-3 T
Therefore, the magnitude of the other magnetic field is 4.39 x 10⁻³ T.

Number of turns in the solenoid, n = 800
Area of cross-section, A =  2.5 × 10^-4 m² 
Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves like a bar magnet because a magnetic field develops along its axis, i.e., along with its length.
The magnetic moment associated with the given current-carrying solenoid is calculated as:
M = n I A
= 800 x 3 x 2.5 x 10^-4
= 0.6 J T^-1

As the electric field is along X-axis, therefore, f
from \(E=-dV/dr\)
(i) dV = -E(dr) = - E(x₂ - x₁)
= - 20(4-0) = -80 V
(ii) dV = -20(\(x_2'-x_1'\))
= -20(6-4)
= -40 V

In a balanced position of the bridge no current will flow in key (K) and hence a constant current will flow in the galvanometer and it will show a constant deflection even when the key K is pressed.

Resistance of the galvanometer coil, G = 15 Ω
Current for which the galvanometer shows full scale deflection,
 = 4 mA = 4 x 10^-3 A
Range of the ammeter is 0, which needs to be converted to 6 A.
Current, I = 6 A
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:
\(S=\frac{I_{g} G}{I-I_{g}}\)
\(=\frac{4 \times 10^{-3} \times 15}{6-4 \times 10^{-3}}\)
\(S=\frac{6 \times 10^{-2}}{6-0.004}=\frac{0.06}{5.996}\)
\(\approx 0.01 \Omega=10 \mathrm{~m} \Omega\)
Hence, a  10 mΩ shunt resistor is to be connected in parallel with the galvanometer.

Energy stored in 6 μF capacitor = E 

\( \therefore \ \frac{1}{2} C_{1} V_{1}^{2}=E \Rightarrow V_{1}^{2}=\frac{2 E}{C_{1}}=\frac{E}{3} \)
\(\because C_{1} \text { and } C_{2} \text { are in parallel, } \therefore V_{1}^{2}=V_{2}^{2}=\frac{E}{3} \)
Now, \(C_{12}=6+12=18 \mu \mathrm{F}\) 
\( \therefore \text { Charge on } C_{12}=q_{12}=C_{12} . V_{12}=18 \times \sqrt{\frac{E}{3}} \)
\(\therefore \text { Charge on } C_{3}=q_{12}=18 \sqrt{\frac{E}{3}} \)
(a) Now, energy stored in 12 μF capacitor
\(=\frac{1}{2} C_{2} V_{1}^{2}=\frac{1}{2} \times 12 \times \frac{E}{3}=2 E\)
(b) Energy stored in 3 μF capacitor
\(=\frac{q_{3}^{2}}{2 C_{3}}=\frac{1}{2} \times\left(\frac{18}{3}\right)^{2} E=18 E\)
\((c) \because C_{e q}=\left(\frac{18 \times 3}{18+3}\right) \mu \mathrm{F}=\frac{18}{7} \mu \mathrm{F}, Q=18 \sqrt{\frac{E}{3}}\)
Energy drawn from the battery,
\(U=\frac{Q^{2}}{2 C_{e q}}=\frac{18 \times 18 \times E \times 7}{3 \times 18}=42 E\)

(a) Expression for the potential energy of a system of two point charges in an external field:
Work done in bringing the charge q₁from infinity to r₁ ,
Work done \(={ q }_{ 1 }V\left( { r }_{ 1 } \right) \)
Work done in bringing the charge qz from infinity to r₂ .
Work done against the external electric field
\(={ q }_{ 2 }V\left( { r }_{ 2 } \right) \)
Work done = work done against the external electric field + Work done on q₂ against the field due to q₁
\(={ q }_{ 2 }V\left( { r }_{ 2 } \right) +\frac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \varepsilon  }_{ 0 }{ r }_{ 12 } } \)
Potential energy of the system = the total work done in assembling the configuration
\(={ q }_{ 1 }V\left( { r }_{ 1 } \right) +{ q }_{ 2 }V\left( { r }_{ 2 } \right) +\frac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \varepsilon }_{ 0 }{ r }_{ 12 } } \)
(b) Electrostatic potential energy of the systems of charges corresponding to initial configuration is
\(U_{i}=\frac{2 k Q^{2}}{l}-\frac{3 k Q^{2}}{l}-\frac{6 k Q^{2}}{l}=-\frac{7 k Q^{2}}{l}\)
Electrostatic potential energy of the system of charges corresponding to final configuration is
\(U_{f}=\frac{4 k Q^{2}}{l}-\frac{6 k Q^{2}}{l}-\frac{12 k Q^{2}}{l}=-\frac{14 k Q^{2}}{l}\)
The amount of work done in shifting the charges to new locations is
\(W=U_{f}-U_{i}\)
\(=-\frac{14 k Q^{2}}{l}-\left(-7 \frac{k Q^{2}}{l}\right)=-\frac{7 k Q^{2}}{l}\)

Number of turns per unit length is given by
\(n=\frac{N}{l}=\frac{100}{4}=25 \mathrm{turns} / \mathrm{m}\)
(i) Magnetic field of a solenoid at a point inside is
\(B=\mu_{0} n i\)
Area of cross-section of the solenoid, A = \(\pi {{r}_{1}}^{2}\) and \( \theta\) = 0° Magnetic flux,
\(\phi_{B}=B A \cos \theta\)
\(=\mu_{0} n i \pi r_{1}^{2} \cos 0^{\circ}\)
\(=4 \pi \times 10^{-7} \times 25 \times 3 \times \pi \times\left(3 \times 10^{-2}\right)^{2}\)
\(=0.27 \times 10^{-6} \mathrm{~Wb}\)
\(=0.27 \mu \mathrm{Wb}\)
(ii) At the end, magnetic field of solenoid is
\(B=\frac{1}{2} \mu_{0} n i\)
\(\therefore \quad \phi_{B}=\frac{0.27}{2}=0.135 \mu \mathrm{Wb}\)

To draw maximum current from a series L-C-R circuit, the circuit at particular frequency
X_L = X_C.
V = \({{1}\over{2\pi\sqrt{LC}}}={{1}\over{2\times314\sqrt{8\times 2\times {10}^{-6}}}}=39.80\) Hz
This frequency is known as the series resonance frequency.
I₀ = \({ { {E}_{0} }\over{ R } }={ {200 }\over{100 } }=2A\)

Sharpness of resonance It is defined as the ratio of the voltage developed across the inductance (L) or capacitance (C) at resonance to the voltage developed across the resistance (R).
Q = \({ { 1 }\over{ R } }=\sqrt{ { { L }\over{ C } } }\)
It may also be defined as the ratio of resonance angular frequency to the bandwidth of the circuit
Q = \({ { {\omega}_{r} }\over{ 2 \triangle \omega } }\)
Circuit become more selective if the resonance is more sharp, maximum current is more, the circuit is close to resonance for smaller range of \((2\triangle \omega)\) of frequencies. Thus, the tuning of the circuit will be good.

(i) A galvanometer of range I_g and resistance G₁ can be converted into
(a) a voltmeter of range V, by connecting a high resistance R in series with galvanometer whose value is given by
\(R=\frac { V }{ { I }_{ g } } -G\)
(b) an ammeter of range I, by connecting a very low resistance (shunt) in parallel with galvanometer whose value is given by
\(S=\frac { { I }_{ g }G }{ I-{ I }_{ g } } -G\)
Thus, the nature of force is attractive.
When direction of flow of current is in opposite direction, the nature of force becomes repulsive.

Since drift velocity \({ v }_{ d }\) and current, I flowing in a conductor are related by the relation:
\({ v }_{ d }=\frac { I }{ neA }.........(1)\) 
\(Also \ drift \ velocity \ in \ terms \ of \ average \ relaxation \ time \ \tau \ is \ given \ by\)
\({ v }_{ d }=\frac { eE\tau }{ m } .........(2)\)
\( From \ (1) \ and \ (2), \ we \ have\)
\( \frac { eE\tau }{ m } =\frac { I }{ neA }\)
\(or\quad \frac { E }{ I } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\( or \ \frac { V }{ lI } =\frac { m }{ n{ e }^{ 2 }A\tau } \)
\( or \ \frac { V }{ I } =\frac { ml }{ n{ e }^{ 2 }A\tau } ...........(3)\)
\( The \ R.H.S. \ of \ Eq(3) \ is \ constant\)
\( \frac { V }{ I } = \ Constant\)
\( This \ is \ Ohm's \ law\)

(i) (a): For stationary electron, \(\vec{v}=0\)
\(\therefore\) Force on the electron is \(\vec{F}_{m}=-e(\vec{v} \times \vec{B})=0\)
(ii) (d): Force on the proton \(\vec{F}_{B}=e(\vec{v} \times \vec{B})\)
Since, \(\vec{v}\) is parallel to \(\vec{B}\)
\(\therefore \quad \vec{F}_{B} \doteq 0\)
Hence proton will continue to move with velocity v along the axis of solenoid.
(iii) (b): Magnetic force on the charged particle q is 
\(\vec{F}_{m}=q(\vec{v} \times \vec{B}) \text { or } F_{m}=q v B \sin \theta\)
where \(\theta\) is the angle between \(\vec{v} \text { and } \vec{B}\)
Out of the given cases, only in case (b) it will experience the force while in other cases it will experience no force
(iv) (a) : \(\vec{F}=q(\vec{v} \times \vec{B})\)
\(=q[(2 \hat{i} \times(\hat{i}+2 \hat{j}+3 \hat{k})]=(4 q) \hat{k}-(6 q) \hat{j}\)
(v) (c): When an electric charge is moving both electric and magnetic fields are produced, whereas a static charge produces only electric field.

(i) (b)
(ii) (a): \(R=\frac{V}{I}=\frac{2}{10^{-6}}=2 \times 10^{6} \Omega\)
(iii) (d): Specific resistance depends upon the nature of material and is independent of mass and dimensions of the material
(iv) (a)
(v) (d): l = 1.0 m; D = 0.4 mm = 4 x 10^-4m
\(R=2 \Omega\)
\(A=\frac{\pi D^{2}}{4}=\frac{\pi \times\left(4 \times 10^{-4}\right)^{2}}{4}=4 \pi \times 10^{-8} \mathrm{~m}^{2}\)
Now, \(\rho=\frac{R A}{l}=\frac{2 \times 4 \pi \times 10^{-8}}{1}=2.55 \times 10^{-7} \Omega \mathrm{m}\)

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
According to Faraday's law of the conservation mechanical energy into electrical energy is in accordance with the law of conservation of energy.

(b): At resonance \(X_{L}=X_{C} \text { or } \omega L=\frac{1}{\omega C}\) Because of this impedance of LCR series circuit become equal to resistance of circuit \(\left(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\right)\).
Therefore from \(I=\frac{E}{Z}=\frac{E}{R},\) at resonance, current in LCR series circuit is maximum. Correspondingly phase angle is also equal to zero. Therefore emf and current are in phase in LCR series circuit.

(c): Consider a conductor of length I and area of cross section A. Time taken by the free electrons to cross the conductor, t = l/v_d.
Hence, current, \(I=\frac{q}{t}=\frac{A l \times n e}{l / v_{d}}\)
or, \(I=\text { Anev }_{d}\)
or, \(I \propto v_{d}\)
Thus current is directly proportional to drift velocity.